Redox reactions drive everything from the rusting of iron to the electricity in your phone battery. In this chapter you will learn to track electrons using oxidation numbers, identify which species is oxidised and which is reduced, write balanced half-reactions, and combine them into a complete, balanced redox equation.
14.1 Oxidation, Reduction, and OIL RIG
The words 'oxidation' and 'reduction' always go together — if one species loses electrons, another must gain them. You cannot have oxidation without reduction occurring at the same time. The mnemonic OIL RIG makes these definitions impossible to forget.
Definition
Oxidation
Oxidation is the loss of electrons by a molecule, atom or ion.
Definition
Reduction
Reduction is the gain of electrons by a molecule, atom or ion.
Note
OIL RIG MNEMONIC: Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). Write 'OIL RIG' in bold at the top of every redox question to remind yourself.
Definition
Redox reaction
A redox reaction is a reaction in which there is a transfer of electrons from one substance to another. Oxidation (loss of electrons) and reduction (gain of electrons) always occur simultaneously — you cannot have one without the other. There is always a change in oxidation numbers of the atoms involved.
In every redox reaction, one species donates electrons (the REDUCING AGENT) and another species accepts those electrons (the OXIDISING AGENT). Note the apparent contradiction: the reducing agent is the species that gets OXIDISED (loses electrons). The oxidising agent is the species that gets REDUCED (gains electrons). This is because they cause the other species to change.
Oxidising Agent vs Reducing Agent
| Property | Oxidising Agent | Reducing Agent |
|---|---|---|
| What it does to itself | Gets REDUCED (gains electrons) | Gets OXIDISED (loses electrons) |
| What it does to the other species | Causes oxidation of the other species | Causes reduction of the other species |
| Change in oxidation number | Oxidation number DECREASES | Oxidation number INCREASES |
| Example (Zn + CuSO₄) | Cu²⁺ (gains 2e⁻ → Cu⁰) | Zn (loses 2e⁻ → Zn²⁺) |
Watch Out
COMMON MISTAKE — Students say 'the oxidising agent is oxidised'. This is WRONG. The oxidising agent causes something else to be oxidised; it itself is REDUCED. Remember: the oxidising agent gains electrons (is reduced). The reducing agent loses electrons (is oxidised).
Worked Example
In the reaction: 2Fe³⁺ + Zn → 2Fe²⁺ + Zn²⁺, identify (a) what is oxidised, (b) what is reduced, (c) the oxidising agent, and (d) the reducing agent.
Given
- Reaction: 2Fe³⁺ + Zn → 2Fe²⁺ + Zn²⁺
Find
Species oxidised, reduced, oxidising agent, and reducing agent.
Solution
- 1Step 1: Track Zn. Zn⁰ → Zn²⁺: charge increases from 0 to +2. It lost 2 electrons → OXIDISED.
- 2Step 2: Track Fe. Fe³⁺ → Fe²⁺: charge decreases from +3 to +2. It gained 1 electron → REDUCED.
- 3Step 3: Zn is oxidised → it is the REDUCING AGENT.
- 4Step 4: Fe³⁺ is reduced → it is the OXIDISING AGENT.
Practice Question
In the reaction Mg + 2HCl → MgCl₂ + H₂, identify: (a) the species oxidised, (b) the species reduced, (c) the oxidising agent, and (d) the reducing agent.
(4 marks)
14.2 Oxidation Numbers
Definition
Oxidation number
The oxidation number (oxidation state) of an atom is the charge that atom would have if the compound were composed entirely of ions. Rules: (1) Oxidation number of a free element = 0. (2) For a simple monatomic ion, oxidation number = ionic charge. (3) In a neutral compound, the sum of oxidation numbers = 0. (4) In a polyatomic ion, the sum = the charge of the ion.
Oxidation numbers are a bookkeeping tool. They allow us to track 'electron ownership' even in covalent compounds where electrons are shared rather than fully transferred. By comparing oxidation numbers before and after a reaction, we can determine which atoms have been oxidised (increase in oxidation number) and which have been reduced (decrease in oxidation number).
Rules for Assigning Oxidation Numbers
- A pure element (uncombined) has an oxidation number of 0. (e.g. Zn⁰, O₂⁰, Fe⁰)
- A monoatomic ion has an oxidation number equal to its charge. (e.g. Na⁺ = +1; Cl⁻ = −1; Fe²⁺ = +2)
- Oxygen in compounds = −2, EXCEPT in peroxides (H₂O₂, Na₂O₂) where O = −1.
- Hydrogen in compounds = +1, EXCEPT in metal hydrides (NaH, CaH₂) where H = −1.
- The sum of all oxidation numbers in a neutral compound = 0.
- The sum of all oxidation numbers in a polyatomic ion = the charge of the ion.
Exam Tip
EXAM TIP — Always apply the rules in order. Start with elements you know (O = −2, H = +1), then use the sum rule to find the unknown oxidation number. Show your working: write the equation 'known + unknown × n = total charge'.
Worked Example
Assign oxidation numbers to all atoms in: (a) KMnO₄ (b) Cr₂O₇²⁻ (c) NH₄⁺
Given
- KMnO₄: neutral compound
- Cr₂O₇²⁻: polyatomic ion with charge 2−
- NH₄⁺: polyatomic ion with charge 1+
Find
Oxidation number of each atom.
Solution
- 1KMnO₄: K = +1 (Group 1). O = −2, so 4 × (−2) = −8. Let Mn = x. +1 + x + (−8) = 0 → x = +7. Mn = +7.
- 2Cr₂O₇²⁻: O = −2, so 7 × (−2) = −14. Let each Cr = x. 2x + (−14) = −2 → 2x = 12 → x = +6. Cr = +6.
- 3NH₄⁺: H = +1, so 4 × (+1) = +4. Let N = x. x + 4 = +1 → x = −3. N = −3.
Worked Example
Is the following reaction a redox reaction? Justify your answer by assigning oxidation numbers: 2Na + Cl₂ → 2NaCl
Given
- Reactants: Na (pure element), Cl₂ (pure element)
- Product: NaCl (ionic compound)
Find
Whether the reaction is a redox reaction.
Solution
- 1Step 1: Na in 2Na: oxidation number = 0 (pure element). Na in NaCl: oxidation number = +1. Change: 0 → +1, INCREASE by 1 → Na is OXIDISED.
- 2Step 2: Cl in Cl₂: oxidation number = 0 (pure element). Cl in NaCl: oxidation number = −1. Change: 0 → −1, DECREASE by 1 → Cl is REDUCED.
- 3Step 3: There are changes in oxidation numbers → the reaction IS a redox reaction.
Note
IDENTIFYING REDOX: A reaction is a redox reaction if and only if at least one element changes its oxidation number. If all oxidation numbers stay the same (e.g. simple acid-base neutralisation where no electron transfer occurs), it is NOT a redox reaction.
Practice Question
Assign the oxidation number of sulfur in each: (a) H₂S (b) SO₂ (c) SO₄²⁻ (d) S₈
(4 marks)
Practice Question
Determine whether the following is a redox reaction and justify: Fe₂O₃ + 3CO → 2Fe + 3CO₂
(5 marks)
14.3 Half-Reactions and Balancing Redox Equations
Definition
Half-reaction
A half reaction is either the oxidation or reduction reaction part of a redox reaction. In the two half-reactions the number of electrons donated equals the number of electrons accepted.
A complete redox reaction can be split into two half-reactions: one showing only the oxidation (electrons released) and one showing only the reduction (electrons consumed). To write a half-reaction: (1) write the species being oxidised or reduced on the left; (2) write the product on the right; (3) add electrons to balance the charge — electrons go on the RIGHT for oxidation and on the LEFT for reduction.
Steps to Balance a Redox Equation Using Half-Reactions
- Write the unbalanced skeletal equation.
- Assign oxidation numbers to identify what is oxidised and what is reduced.
- Write the oxidation half-reaction (species losing electrons). Balance atoms, then add e⁻ on the RIGHT.
- Write the reduction half-reaction (species gaining electrons). Balance atoms, then add e⁻ on the LEFT.
- Multiply each half-reaction by a factor so the number of electrons in both half-reactions is equal.
- Add the two half-reactions together and cancel species that appear on both sides (including electrons).
- Check: all atoms balanced and all charges balanced.
Worked Example
Balance the following using half-reactions: Zn + CuSO₄ → ZnSO₄ + Cu
Given
- Ionic form: Zn + Cu²⁺ → Zn²⁺ + Cu (SO₄²⁻ is spectator ion)
Find
Balanced overall redox equation and the two half-reactions.
Solution
- 1Step 1: Oxidation numbers. Zn: 0 → +2 (increases → oxidised). Cu: +2 → 0 (decreases → reduced).
- 2Step 2: Oxidation half-reaction. Zn → Zn²⁺ + 2e⁻ (Zn loses 2 electrons)
- 3Step 3: Reduction half-reaction. Cu²⁺ + 2e⁻ → Cu (Cu²⁺ gains 2 electrons)
- 4Step 4: Both half-reactions already have 2 electrons — no scaling needed.
- 5Step 5: Add: Zn + Cu²⁺ + 2e⁻ → Zn²⁺ + 2e⁻ + Cu. Cancel 2e⁻ from both sides.
- 6Step 6: Overall ionic equation: Zn + Cu²⁺ → Zn²⁺ + Cu. Full equation: Zn + CuSO₄ → ZnSO₄ + Cu.
Worked Example
Balance the redox reaction between MnO₄⁻ and Fe²⁺ in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic solution)
Given
- MnO₄⁻ is reduced (Mn: +7 → +2, gains 5e⁻)
- Fe²⁺ is oxidised (Fe: +2 → +3, loses 1e⁻)
- Reaction occurs in acidic solution — H⁺ and H₂O can be used to balance
Find
Balanced half-reactions and overall equation.
Solution
- 1Step 1: Reduction half-reaction. MnO₄⁻ → Mn²⁺. Balance O: add 4H₂O on right. Balance H: add 8H⁺ on left. Balance charge: left = 8(+1) + (−1) = +7; right = +2; add 5e⁻ on left: 8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O. ✓
- 2Step 2: Oxidation half-reaction. Fe²⁺ → Fe³⁺ + e⁻. ✓
- 3Step 3: Scale so electrons cancel. Multiply Fe half-reaction by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻.
- 4Step 4: Add: 8H⁺ + MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Electrons cancel. ✓
Watch Out
COMMON MISTAKE — Forgetting to balance charge when writing half-reactions. After balancing all atoms, count the total charge on each side. The difference is the number of electrons to add. Electrons are ALWAYS added to the more positive (or less negative) side.
Exam Tip
EXAM TIP — In Grade 11, most half-reactions involve simple electron transfer without oxygen atoms (e.g. Zn/Cu, Fe²⁺/Fe³⁺). Focus on getting these perfect before tackling MnO₄⁻ type reactions. Always check your final equation: total charge on left must equal total charge on right, AND all atoms must balance.
Practice Question
Write balanced half-reactions for: (a) Al → Al³⁺ (oxidation) (b) Ag⁺ → Ag (reduction). Then combine them to write the overall balanced equation.
(6 marks)
Practice Question
Given: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (reduction) and Fe²⁺ → Fe³⁺ + e⁻ (oxidation). Combine to give the balanced overall equation.
(4 marks)
Practice Question
In the reaction Cl₂ + 2KBr → 2KCl + Br₂: (a) Assign oxidation numbers to all atoms. (b) Identify the oxidising agent and reducing agent. (c) Write balanced half-reactions.
(6 marks)